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Proof step 6 from proof bag-rm-add-eq (previous step)

(count(a, bag) - (count(a, bag) ≠ 0 ⊃ 1;0)) + 1 = count(a, bag) ⊦ a ∈ bag

User applied rule case distinction

And obtained the following new goals

Proof step 7:
(count(a, bag) - 1) + 1 = count(a, bag) ⊦ count(a, bag) = 0, a ∈ bag

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ m - 0 = m; (s)% ⊦ m = m + n ↔ n = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic